Question

1 a 0.7 kg sphere is placed on top of a coiled spring. the coiled spring has 32 j of elastic potential energy. when the spring is released, the sphere moves upward, a n in the diagram below. neglecting friction and the mass of the what is the change in height (δh) of the sphere? a 224.0 m b 4.5714285714 m c 45.7142857143 m d 0.021875 m

Explanation:

Step1: Apply energy conservation

The elastic potential energy of the spring \(E_{p - spring}=32\space J\) is converted into gravitational potential energy of the sphere \(E_{p - gravity}=mgh\). So, \(E_{p - spring}=E_{p - gravity}\), i.e., \(32 = mgh\).

Step2: Solve for \(\Delta h\)

Given \(m = 0.7\space kg\) and \(g = 9.8\space m/s^{2}\). From \(h=\frac{E_{p - spring}}{mg}\), substitute the values: \(h=\frac{32}{0.7\times9.8}\).
Calculate \(0.7\times9.8 = 6.86\), then \(h=\frac{32}{6.86}\approx4.5714285714\space m\).

Answer:

B. \(4.5714285714\space m\)